# Conductivity from dielectric permittivity

If I have a typical Drude dielectric permittivity:

$$\epsilon(\omega) = \epsilon_{\infty} - \frac{\omega_p^2 \tau}{\omega^2\tau + i\omega}$$

Now, decomposing $$\Re{(\epsilon(\omega))}$$ and $$\Im{(\epsilon(\omega))}$$:

$$\Re{(\epsilon(\omega))}=-\frac{\omega_p^2}{\omega^2+1/\tau^2} + \epsilon_\infty$$ $$\Im{(\epsilon(\omega))} = \frac{\omega^2_p}{\omega^3 \tau+1/\tau}$$

Now, the imaginary part is the important one, because it relates the permittivity to conductivity (which is what I want).

$$\epsilon_{\text{im}}(\omega)= \frac{\sigma(\omega)}{\omega \epsilon_0}$$

So, why does equating the imaginary part and the above equation not yield the correct answer?

$$\sigma(\omega)=\frac{\omega_p^2\epsilon_0}{\omega^2\tau+1/\tau}$$

Should be:

$$\sigma_D(\omega) = \frac{\epsilon_0\omega_p^2 \tau}{1-i\omega\tau}$$

Where am I going wrong?

You’re going wrong by considering the optical conductivity to be real only. Conductivity is complex just as permittivity is! It can be related to epsilon through Ampere’s Law: $$\nabla\times H = J + \frac{dD}{dt}$$ For time-harmonic fields, we get $$\nabla\times H = (\sigma - i \omega \epsilon)E.$$ So the complex relative permittivity is related to the complex conductivity as $$\epsilon_r=\frac{\sigma}{i \omega \epsilon_0} - 1.$$ This is general (although there might be a sign discrepancy depending on your phasor convention) and, depending on your mood, can be taken as a matter of definition. If you leave out the imaginary part of $$\sigma$$, then you lose half of the information contained in the Drude model!
• @smollma Try replacing the 1 in my expression with $\epsilon_\infty$ and possibly taking the complex conjugate. Let me know how it goes! Nov 3 '20 at 13:03
• Yes, I understand that the conductivity should be complex. I was trying to equate the imaginary part of the conductivtiy and imag. part of permittivity (ignoring $i$) to use this to swap between the two functions. I can't seem to do it though. So: $$\epsilon_{\infty} +i \frac{\sigma(\omega)}{\omega \epsilon_0} = \epsilon_{\infty} - \frac{\omega_p^2 \tau}{\omega^2 \tau +i\omega}$$ This leaves me with: $$\sigma(\omega) = -i 2 \epsilon_{\infty}\omega\epsilon_0 - \frac{\omega_p^2\omega\epsilon_0\tau}{j(\omega^2\tau +i\omega)}$$ Assume $\epsilon_\infty$ is 1? Nov 5 '20 at 17:25